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Source. Section 6, preprint pp. 11--12: the introductory paragraph, Theorem 6.1 and its proof, and Theorem 6.2. Read on the rendered pages.

Statement

Theorem 6.1 (p. 11): "Suppose that $\{a_n\}_{n=1}^{\infty}$ is a monotonic sequence of positive integers such that $\lim_{n\to\infty}\frac{a_n}{\log n}=\infty$. Then $S=\sum_{n=1}^{\infty}\frac{p_n}{a_1\ldots a_n}$ is rational if and only if $\frac{p_n}{a_n-1}$ is constant for $n\ge n_0$."

Section 6 opens (p. 11) by relaxing the condition $p_n=o(a_n^2)$ of Theorem 5.1, which amounts to $a_n/\sqrt{n\log n}\to\infty$, to $a_n/\log n\to\infty$, and goes on: "In this way we partially affirm the expectation expressed by Erdős in [3] p.99 that only the monotonicity of $\{a_n\}_{n=1}^{\infty}$ suffices. We are not able to prove the irrationality of $\sum_{n=1}^{\infty}\frac{p_n}{2^n}$ either."

Proof structure (pp. 11--12)

Suppose $S=r/q$. Since $p_n\le2n\log n$ and $p_{n+1}\le1.1p_n$ for large $n$, the tail of $S_n$ beyond $K_n=2[\log n]+1$ terms is less than $1/(8q)$. By Theorem 5.1 one may assume infinitely many $N$ with $a_{2N}<N^{0.6}$; for such $N$, at most $N^{0.6}$ indices in $[N,2N)$ have $a_{n+1}>a_n$, and with $K=K_{2N}\le2.1\log N$ and $p_{2N}<4N\log N$, the gap $p_{n+1}-p_n>20(\log N)^2$ occurs for at most $N/(5\log N)$ indices; so a set $A$ of more than $N/2$ indices $n\in[N,2N)$ has $a_n=\cdots=a_{n+K}$ and all gaps $p_{n+i+1}-p_{n+i}\le20(\log N)^2$, $i\le K$. For $n\in A$, (6) bounds $|(S_n-S_{n+1})-(p_n/a_n-p_{n+1}/a_{n+1})|$ by $1/(2q)$. Then $S_n=S_{n+1}$ would give $p_n\mid(a_n-1)$, impossible as $a_n<n^{0.6}$; $S_{n+1}<S_n$ is impossible by (6) and $q(S_n-S_{n+1})\in\mathbb{Z}$; and $S_{n+1}>S_n$ for all $n\in A$ gives $p_{n+1}-p_n\ge a_N/(2q)>10\log N$ on a set of size at least $N/2$, so $p_{2N}>5N\log N$, contradicting $p_{2N}<4N\log N$. $\blacksquare$

Theorem 6.2 (p. 12) is the analog with $b_n=n$: for an unbounded monotonic sequence of positive integers $a_n$, $\sum n/(a_1\ldots a_n)$ is rational if and only if $n/(a_n-1)$ is constant for $n\ge n_0$; for bounded monotonic $a_n$ the sum is rational.

Relation to problem 251

The hypothesis $a_n/\log n\to\infty$ excludes every bounded sequence, in particular $a_n=2$, so problem 251 is untouched, as the authors say. Its growth condition on monotonic $a_n$ is weaker than those of the earlier results on the same series: Erdős 1958 needed $q_n(\log n)^k/n\to\infty$, Erdős–Straus 1974 and Theorem 5.1 need $p_n=o(a_n^2)$. Monotonicity cannot be dropped (Remark on pp. 9--10, and the non-monotone counterexample claimed in the 2026 Kovač note against the 1988 expectation).

Bears on. #251 (the monotone relatives of the problem's series; the problem's own case is excluded and declared out of reach on p. 11).